Sketch the graphs and set up an integral for the area. The strip whose area is da would be a rectangle whose height is the distance between the functions. The width is dx. Integrate over the limits you already have determined.

Sketch the graphs and set up an integral for the area. The strip whose area is da would be a rectangle whose height is the distance between the functions. The width is dx. Integrate over the limits you already have determined.

You need to sketch the graphs of those two curves to see the limits of integration.

The points of intersection are: (-1,0) and (5,6).

In this case, there are 2 ways of finding the area between those 2 curves. The simplest way is to describe the enclosed region like this:
For x fixed, y varies from y=... to y=...
x varies from x=... to x=...

Despite what some people are saying, you don't need to sketch the graphs, though it is a nice sanity check of your answer.

You said you already have the points of intersection. The next step is to determine which function is on top. To do this, you can evaluate each function at a point in the middle of your 2 points of intersection and see which function is greater. (If you insist on doing it without graphing). This method is also good if you are dealing with functions whose shapes are not immediately obvious. Since in this problem the shapes are pretty particular, you can also do it by graphing the two functions and examining the region of interest.

You use integration to get the area under the top curve and subtract it by the area under the lower curve (using integration to find it too). This leaves the area in the middle of the two curves. HallsofIvy details this process. If you look at the graph above me, this can be visualized as "total area under the rhombus that is made from that linear line all the way down to the x-axis minus the total area under the U shaped curve all the way down to the x-axis (all between the 2 points of interest -- the two points of intersection)" That leaves the area of the figure that is sort of D shaped and rotated to the right a bit (the are between the curves).

One other thing. You do not "solve" an integral - you evaluate it. When you solve an equation (for some variable), you rewrite the equation in a form in which <some variable> is isolated on one side of the equation.

I have a question about how to present your solution to questions like this, my question is trivial so I'll post here first and if no one answers I will post a new thread.

After finding the intercept points between the 2 functions, the equation you need to solve looks something like this...

area = ∫(x+1)-(x^2-3x-4) dx

I am wondering why there is only one "dx", to me this looks like it's saying we are only going to find the integral of (x^2-3x-4) and not (x+1).

Shouldn't it be presented like this?

area = ∫((x+1)-(x^2-3x-4)) dx

or this?

area = ∫(x+1) dx -(x^2-3x-4) dx

What am I missing here?

When you think about the properties of integrals, it is useful to consider a summation that approximates it since it is less mysterious. Consider this:
[tex]\sum(x+1)h-\sum(x^2-3x-4)h[/tex]
where we are summing small rectangles of width h and height equal to the functions evaluated at particular values of x (i.e. x+1 and x^2-3x-4). The summation would go from the lower limit to the upper limit with a step size of h (e.g. if h = .01 and the integral was from 0 to 1, you would sum up the values at x = 0, x = .01, etc.)

So we can work with these more intuitive summations to show how it equals the thing you are confused about it equaling. The sum of a plus the sum of b is the sum of (a+b). Are you comfortable with that? If so, we get

[tex]\sum[(x+1)h-(x^2-3x-4)h][/tex]

Then, h is just a number, so we can factor it out.

[tex]\sum[(x+1)-(x^2-3x-4)]h[/tex]

Then (a+b) - (c-d-e) = a+b-c+d+e...
[tex]\sum[x+1-x^2+3x+4]h[/tex]

Since an integral is basically this type of summation with h = dx as h goes toward 0, it is also true that

[tex]\int x+1 - x^2 +3x+4 dx=\int (x+1) dx - \int (x^2-3x-4) dx[/tex]

But since it makes no sense to have a term inside an integral that isn't multiplied by your dx, people just leave off that internal parentheses:
[tex]\int (x+1) dx - \int (x^2-3x-4) dx=\int x+1 dx - \int x^2-3x-4 dx[/tex]

Using this interpretation, I also see no problem with your writing of

∫(x+1) dx -(x^2-3x-4) dx

as long as we put another bracket to indicate the infinite summation is over both portions

You do not "solve" an integral - you evaluate it. When you solve an equation (for some variable), you rewrite the equation in a form in which <some variable> is isolated on one side of the equation.

Thanks, I intend to leave no stone unturned in coming to understand every technical detail in A-level mathematics, I have covered "pure 1" so far and I will ask a bunch of well presented questions like this again tomorrow after going through "pure 2", all of which I will post in one new thread.